bulbareeh5kl

2022-06-03

How do I prove $B=C$ when...
If $A\cap B=A\cap C$ and ${A}^{\complement }\cap C$, then $B=C$.
I seem to be able to prove this without the use of ${A}^{\complement }\cap C$ but it must be there for a reason. Where would I add this in. The basic gist behind what I did was say x was an element of A therefore x is in $A\cap B$ and $A\cap C$. So x is in B and x is in C. So since they share the same element they are equal. This may not be right but I think I'm somewhere in the ballpark. Any help would be greatly appreciated.

polomierzvxe4b

Step 1
The basic gist behind what I did was say x was an element of A therefore x is in $A\cap B$ and $A\cap C$ (though that doesn't really help you with this proof)
Here is a mistake. Just because $x\in A$ does not mean $x\in A\cap B$, because for the latter to be true, it needs to also be true that $x\in B$, and you don't know that. What you can say is that if $x\in A$, then $x\in A\cup B$ and $x\in A\cup C$
Step 2
So x is in B and x is in C. So since they share the same element they are equal.
This is another mistake: even if your earlier reasoning went through, all you have shown that all elements that are in A are in both B and C. This does not rule out any objects that might be in B but not in C, and that would be possible for objects that are not in A in the first place. But note: the second given that ${A}^{C}\cap B={A}^{C}\cap C$ will help you rule out exactly this latter possibility.

Trystan Yates

Step 1
Since the problem is symmetric for B and C it is enough to show that $B\subset C$. Let x be an element of B. If x is in A then it is also in C since $A\cap B=A\cap C$. If x is not in A then it is also in C since $x\in {A}^{C}\cap B={A}^{C}\cap C$.
Step 2
In any case, $x\in B\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x\in C$ so $B\subset C$.
Your proof was wrong because you assumed two subsets sharing an element are the same, which is false. "Countries in Europe" and "Countries in the world" share many elements but are obviously not the same set.

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